Problem: A $39$ -meter ladder is sliding down a vertical wall so the distance between the bottom of the ladder and the wall is increasing at $10$ meters per minute. At a certain instant, the bottom of the ladder is $36$ meters from the wall. What is the rate of change of the distance between the top of the ladder and the ground at that instant (in meters per minute)? Choose 1 answer: Choose 1 answer: (Choice A) A $-\dfrac{25}{6}$ (Choice B) B $-10$ (Choice C) C $-12$ (Choice D) D $-24$
Setting up the math Let... $a(t)$ denote the distance between the top of the ladder and the ground at time $t$, $b(t)$ denote the distance between the bottom of the ladder and the wall at time $t$, and $c$ denote the length of the ladder (which is always $39$ meters). $a(t)$ $b(t)$ $c$ We are given that $c=39$ and $b'(t)=10$. We are also given that $b(t_0)=36$ for a specific time $t_0$. We want to find $a'(t_0)$. Relating the measures The measures relate to each other through the Pythagorean theorem: $\begin{aligned} [a(t)]^2+[b(t)]^2&=c^2 \\\\\\ [a(t)]^2+[b(t)]^2&=39^2 \end{aligned}$ We can differentiate both sides to find an expression for $a'(t)$ : $a'(t)=-\dfrac{b(t)b'(t)}{a(t)}$ Using the information to solve In order to find $a'(t_0)$ we need to find $a(t_0)$. Using the Pythagorean theorem and the fact that $b(t_0)=36$ and $c=39$, we can find that $a(t_0)=15$. Let's plug ${b(t_0)}={36}$, ${b'(t_0)}={10}$, and ${a(t_0)}={15}$ into the expression for $a'(t_0)$ : $\begin{aligned} a'(t_0)&=-\dfrac{{b(t_0)}{b'(t_0)}}{{a(t_0)}} \\\\ &=-\dfrac{({36})({10})}{({15})} \\\\ &=-24 \end{aligned}$ In conclusion, the rate of change of the distance between the top of the ladder and the ground at that instant is $-24$ meters per minute. Since the rate of change is negative, we know that the distance is decreasing.